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Snezhnost [94]
3 years ago
14

** In Problem 4.26, the coefficient of static friction between the book and the vertical back of the wagon is μs. Determine an e

xpression for the minimum acceleration of the wagon in terms of μs so that the book does not slide down. Does the mass of the book matter? Explain.
Physics
1 answer:
tangare [24]3 years ago
7 0

Answer:

Explanation:

The situation is that a wagon is speeding up with a book remaining stuck with the vertical wall of the wagon . Due to acceleration of wagon , book pushes the vertical wall of wagon with some force .

Minimum acceleration be a . Acceleration of book will be a. Force on book will be ma . Reaction force R of vertical back of the wagon will be ma so friction force f  generated on the book in upward  direction will be μs x R

This friction force f will balance the weight . So

mg =  μs x R

= μs x ma

a = g / μs

Mass has no effect on this acceleration.

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Set the radius to 2.0 m and the velocity to 1.0 m/s. Keeping the radius the same, record the magnitude of centripetal accelerati
jek_recluse [69]

Answer:

a=4\ m/s^2

Explanation:

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3 years ago
The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N
guajiro [1.7K]

Answer:

\frac{r_1}{r_2}=6.9

Explanation:

According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:

P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}

where,

F₁ = Load lifted by output plunger = 2100 N

F₂ = Force applied on input piston = 44 N

r₁ = radius of output plunger

r₂ = radius of input piston

Therefore,

\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9

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3 years ago
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